1,1

Nathaniel Johnston, Table of n, a(n) for n = 1..10000

A010888(a(n))=A010888(b(n)) where a(n)=A001358(k), b(n)=A001358(k+1). - R. J. Mathar, May 23 2006

146 and 155 are two consecutive semiprimes with the same digital root, namely 2.

with(numtheory): A001358 := proc(n) option remember: local k: if(n=1)then return 4:fi: for k from procname(n-1)+1 do if bigomega(k) = 2 then return k: fi: od: end: A118699ind := proc(n) option remember: local k: if(n=1)then return 50:fi: for k from procname(n-1)+1 do if(A001358(k) mod 9 = A001358(k+1) mod 9)then return k: fi: od: end: seq(A001358(A118699ind(n)), n=1..20); # Nathaniel Johnston, May 04 2011

Select[Partition[Select[Range[6400], PrimeOmega[#]==2&], 2, 1], 1+Mod[#[[1]]-1, 9] == 1+Mod[#[[2]]-1, 9]&][[All, 1]] (* Harvey P. Dale, Oct 23 2019 *)

(PARI) isA001358(n)= { local(f, fct, sumpo); if(n <4, return(0) ); f=factor(n); fct=(matsize(f))[1]; sumpo= sum(i=1, fct, f[i, 2]); if ( sumpo != 2, return(0), return(1) ); } A010888(n)= { if( n %9 != 0, return(n % 9), return(9) ); } { oldr=0; oldn=0; for(n=5, 10000, if( isA001358(n)==1, dr=A010888(n); if(dr==oldr, print1(oldn, ", "); ); oldr=dr; oldn=n; ); ); } \\ R. J. Mathar, May 23 2006

Cf. A001358.

Sequence in context: A135666 A365202 A119379 * A238579 A238028 A179572

Adjacent sequences: A118696 A118697 A118698 * A118700 A118701 A118702

base,easy,nonn

Luc Stevens (lms022(AT)yahoo.com), May 20 2006

Corrected and extended by R. J. Mathar, May 23 2006

approved